1044 Shopping in Mars （25 分)

2019-03-19

1044 Shopping in Mars （25 分)

Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken off the chain one by one. Once a diamond is off the chain, it cannot be taken back. For example, if we have a chain of 8 diamonds with values M$3, 2, 1, 5, 4, 6, 8, 7, and we must pay M$15. We may have 3 options:

Cut the chain between 4 and 6, and take off the diamonds from the position 1 to 5 (with values 3+2+1+5+4=15).
Cut before 5 or after 6, and take off the diamonds from the position 4 to 6 (with values 5+4+6=15).
Cut before 8, and take off the diamonds from the position 7 to 8 (with values 8+7=15).

Now given the chain of diamond values and the amount that a customer has to pay, you are supposed to list all the paying options for the customer.

If it is impossible to pay the exact amount, you must suggest solutions with minimum lost.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 numbers: N (≤105), the total number of diamonds on the chain, and M (≤108), the amount that the customer has to pay. Then the next line contains N positive numbers D1⋯D**N (D**i≤103 for all i=1,⋯,N) which are the values of the diamonds. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print i-j in a line for each pair of i ≤ j such that Di + … + Dj = M. Note that if there are more than one solution, all the solutions must be printed in increasing order of i.

If there is no solution, output i-j for pairs of i ≤ j such that Di + … + Dj >M with (Di + … + Dj −M) minimized. Again all the solutions must be printed in increasing order of i.

It is guaranteed that the total value of diamonds is sufficient to pay the given amount.

Sample Input 1:

1 2	16 15 3 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13

Sample Output 1:

1-5
4-6
7-8
11-11

Sample Input 2:

1 2	5 13 2 4 5 7 9

Sample Output 2:

1
2

2-4
4-5

作者: CHEN, Yue

单位: 浙江大学

时间限制: 300 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

从给定的一组数据中，求连续数之和等于给定数的区间。如果没有相应的区间，那么输出序列和大于给定数的最小值的区间。

分析

暴力易于实现，但是有两个测试点超时。网上有一种o(n)的解法，比较巧妙。思路如下：

维持一个low起点与i终点，读入i处的数。如果low与i之间的和已经超过m，那么前进low，从而减小区间和，判断和是否为m。如果和小于m，显然，i++，继续读入。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include<sstream>

using namespace std;

int main() {
    int num[100005];
    int n,m,minsum=99999999,flag=0;
    int sum[100005];
    map<int,int> mincut;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        cin>>num[i];
    int low=0;
    sum[0]=0;
    for(int i=1;i<=n;i++){
        sum[i]=sum[i-1]+num[i];
        while(low<i&&sum[i]-sum[low]>m){
            //记录大于m的最小值
            if(sum[i]-sum[low]<minsum){
                mincut.clear();
                mincut[low+1]=i;
                minsum=sum[i]-sum[low];
            } else if(sum[i]-sum[low]==minsum)
                mincut[low+1]=i;
            low++;
        }
        if(sum[i]-sum[low]==m) {
            flag=1;
            cout << low + 1 << "-" << i << endl;
        }
    }
    if(!flag){
        for(auto it=mincut.begin();it!=mincut.end();it++){
            cout<<it->first<<"-"<<it->second<<endl;
        }
    }
    return 0;
}